Example 16 - If A, B, C are three events associated with a random - Using formulae of sets

Example  16 - Chapter 16 Class 11 Probability - Part 2

This video is only available for Teachoo black users

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Example 11 If A, B, C are three events associated with a random experiment, prove that P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(A ∩ B) − P(A ∩ C) – P (B ∩ C) + P ( A ∩ B ∩ C) Let B ∪ C = E So, P (A ∪ B ∪ C) = P(A ∪ E) = P(A) + P(E) – P(A ∩ E) = P(A) + P(E) – P(A ∩ (B ∪ C)) = P(A) + P(E) – P((A ∩ B) ∪ (A ∩ C)) We find P(E) & P((A ∩ B) ∪ (A ∩ C)) separately Finding P(E) P (E) = P (B ∪ C) = P(B) + P(C) − P(B ∩ C) Finding P((A ∩ B) ∪ (A ∩ C)) We know that P(A ∪ B) = P(A) + P(B) – P(A ∩ B) Putting A = (A ∩ B) & B = (A ∩ C) P((A ∩ B) ∪ (A ∩ C)) = P(A ∩ B) + P(A ∩ C) – P ((A ∩ B) ∩ (A ∩ C)) P((A ∩ B) ∪ (A ∩ C)) = P(A ∩ B) + P(A ∩ C) – P (A ∩ B ∩ C) Putting (2) & (3) in (1) P (A ∪ B ∪ C) = P(A) + P(E) – P((A ∩ B) ∪ (A ∩ C)) = P(A) + [P(B) + P(C) – P(B ∩ C)] – [P(A ∩ B) + P(A ∩ C) + P(A ∩ B ∩ C)] = P(A) + P(B) + P(C) − P(A ∩ B) − P(A ∩ C) – P (B ∩ C) + P ( A ∩ B ∩ C) Hence proved

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.